If the roots of the equation x^2 - 8x + (a^2 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The roots of the quadratic equation $x^2 - 8x + (a^2 - 6a) = 0$ are real if and only if the discriminant $D \ge 0$.For the equation $Ax^2 + Bx + C

Vedclass · Accepted Answer

$ - 2 \le a \le 8$

Vedclass · Answer

(C) The roots of the quadratic equation $x^2 - 8x + (a^2 - 6a) = 0$ are real if and only if the discriminant $D \ge 0$.For the equation $Ax^2 + Bx + C = 0$,$D = B^2 - 4AC$.Here,$A = 1$,$B = -8$,and $C = a^2 - 6a$.$D = (-8)^2 - 4(1)(a^2 - 6a) \ge 0$$64 - 4(a^2 - 6a) \ge 0$$16 - (a^2 - 6a) \ge 0$$16 - a^2 + 6a \ge 0$$a^2 - 6a - 16 \le 0$Factoring the quadratic expression:$(a - 8)(a + 2) \le 0$For the product of two factors to be less than or equal to zero,the value of $a$ must lie between the roots of the equation $(a - 8)(a + 2) = 0$,which are $a = 8$ and $a = -2$.Thus,the inequality $(a - 8)(a + 2) \le 0$ holds for $a \in [-2, 8]$.Therefore,$-2 \le a \le 8$.

If the roots of the equation $x^2 - 8x + (a^2 - 6a) = 0$ are real,then:

Similar Questions

Given the standard electrode potentials,
$K^{+} / K = -2.93 \ V$,$Ag^{+} / Ag = 0.80 \ V$
$Hg^{2+} / Hg = 0.79 \ V$
$Mg^{2+} / Mg = -2.37 \ V$,$Cr^{3+} / Cr = -0.74 \ V$
Arrange these metals in their increasing order of reducing power.

In terms of period and group,where would you locate the element with $Z=114$?

If the eccentricity of a hyperbola is $\frac{5}{3}$,then the eccentricity of its conjugate hyperbola is

If $(a+bx)^{-3} = \frac{1}{27} + \frac{1}{3}x + \dots$,then the ordered pair $(a, b)$ equals to

The process of the isolation of a metal by dissolving the ore in a suitable chemical reagent followed by precipitation of the metal by a more electropositive metal is called:

Vedclass Test Series

Exam Paper Generator

Online Exam Module